{c}

Let $\{U_i\}$ be an open cover of $f(X)$. By <continuity> the inverse images $f^{-1}(U_i)$ are open and cover $X$. Compactness provides finitely many of them covering $X$, and the corresponding $U_i$ then cover $f(X)$. Thus \b[a continuous image of a <compact> space is <compact>].

Solution

Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / E / 2 / a / ii / Solution

 Solution (source code)